int_0^{frac{pi}{2}} sin^6 x cos^4 x , dx = | vedclass

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(C) Let $I = \int_0^{\frac{\pi}{2}} \sin^6 x \cos^4 x \, dx$. Using Wallis's formula, $\int_0^{\frac{\pi}{2}} \sin^m x \cos^n x \, dx = \frac{(m-1)(m-

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$\frac{3\pi}{512}$

Vedclass · Answer

(C) Let $I = \int_0^{\frac{\pi}{2}} \sin^6 x \cos^4 x \, dx$. Using Wallis's formula, $\int_0^{\frac{\pi}{2}} \sin^m x \cos^n x \, dx = \frac{(m-1)(m-3)\dots(1) 	imes (n-1)(n-3)\dots(1)}{(m+n)(m+n-2)\dots(2)} 	imes \frac{\pi}{2}$ (if both $m, n$ are even). Here, $m = 6$ and $n = 4$. $I = \frac{(6-1)(6-3)(6-5) 	imes (4-1)(4-3)}{(6+4)(6+4-2)(6+4-4)(6+4-6)(6+4-8)} 	imes \frac{\pi}{2}$ $I = \frac{5 	imes 3 	imes 1 	imes 3 	imes 1}{10 	imes 8 	imes 6 	imes 4 	imes 2} 	imes \frac{\pi}{2}$ $I = \frac{45}{3840} 	imes \frac{\pi}{2} = \frac{3}{256} 	imes \frac{\pi}{2} = \frac{3\pi}{512}$.

$\int_0^{\frac{\pi}{2}} \sin^6 x \cos^4 x \, dx =$

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